Formula question

[Fishing Folks]

I need to know if there is an algebraic formula for finding the area of
an octagon. Please e-mail me if you can help...

Thanks,
Chuck

[Ktulwxwatcher]

Yes, there is a formula that says the area of a regular polygon is A=1/2ap
where a is the apothem (the perpendicular distance to a side from the center)
and p is the perimeter. If you need help finding the apothem, send me the
problem and I can work it and show you how I did it.

David

[Stan Brown]

[cc'd to previous poster for speed; please follow up in newsgroup]

Fishing Folks <ste...@webtv.net> wrote in alt.algebra.help:

>I need to know if there is an algebraic formula for finding the area of
>an octagon.

Of course there is -- several formulas in fact.

If you mean _any_ octagon, the easiest thing to do is probably draw
lines from vertices to center to break it into eight triangles, and
sum the areas of those triangles using good old base times height
over 2.

If you mean a _regular_ octagon, one with all eight sides equal,
then it's very much easier. You could look up a formula at
<http://www.geom.umn.edu/docs/reference/CRC-formulas/>, but where's
the fun in that?[1] Instead, let's derive the formula.

(Why derive it instead of looking it up? If not for the fun of it --
and, strangely, some people don't think of it as fun -- remember
that on an exam you will not have access to Internet resources, but
you will still have your own brain. Learn to derive things at need,
and you will be able to solve more problems at need.)

A regular octagon is a square with the four corners cut out, each
cut-out corner being a right triangle, half of a small square. We
know how to find the area of a square and the areas of triangles, so
it's just a subtraction.

If the side of the octagon is s, then s is also the diagonal of each
cutout, and the side of each cutout is s/sqrt(2). This means the
side of the original big square is the side of the octagon plus the
sides of the two cutouts, s+2s/sqrt(2) = s + s*sqrt(2), and its area
is
A1 = [s+s*sqrt(2)]^2
= s^2 * [1 + sqrt(2)]^2
= s^2 * (1 + 2*sqrt(2) + 2)
= s^2 * (3 + 2*sqrt(2))
Now each of the triangular cutouts has area
A2 = (1/2)[s/sqrt(2)]^2
= (1/2) s^2 / 2
= (1/4)s^2
and since there are four cutouts the area of the octagon must be the
area A1 of the original square minus the areas 4*(A2) of the four
cutouts:
A = A1 - 4*(A2)
= s^2 * (3 + 2*sqrt(2)) - 4*(1/4)*s^2
= s^2 * (3 + 2*sqrt(2) - 1)
= s^2 * (2 + 2*sqrt(2))
= 2(1+sqrt(2))s^2
= about 4.83s^2.

>Please e-mail me if you can help...

Please read "Just e-mail me the answer!" in the FAQ.

[1] As it happens, there's an error on that page. The formula given,
for a regular polygon of n sides each a long, is
area = (1/4)(ks^2)cot(180°/k)
where s = the half-perimeter, ka/2. It should be
area = (1/4)(ka^2)cot(180°/k)
(The error is easy to miss if you assume s is the length of a side
instead of the half-perimeter it is defined to be.) For a regular
octagon, the area is
area = 2(a^2)cot(22.5°)
= 2(1+sqrt(2))a^2
and the given formula yields a result 16 times the correct result.

--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://oakroadsystems.com
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